# Issue with integral in dimensional regularization

So I am calculating the renormalization group rquations for some exotic new particles and used dimensional regularization for all the calculations up to this point. Now I am look at the vertex corrections in the massless limit in which all external momentum are equal to 0. The issue here is I end up with an integral I don't know how to solve.

Its this integral
$$\int d^{4}l \frac{1}{l^4}$$
which you obviously can't solve using dimensional regularization. What other method can I use to solve it and how will I make the solution compatable with my other solutions since they will have the ##\frac{1}{\epsilon}##

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BiGyElLoWhAt
Gold Member
Well, I don't know much about renormalization, but that integral should be no different than ##\int \frac{1}{\ell^4}dl*dl*dl*dl##
I drew that from the ##d\tau## notation that means ##\int d\tau = \int\int\int dx*dy*dz## Does that help you solve it?

stevendaryl
Staff Emeritus
So I am calculating the renormalization group rquations for some exotic new particles and used dimensional regularization for all the calculations up to this point. Now I am look at the vertex corrections in the massless limit in which all external momentum are equal to 0. The issue here is I end up with an integral I don't know how to solve.

Its this integral
$$\int d^{4}l \frac{1}{l^4}$$
which you obviously can't solve using dimensional regularization. What other method can I use to solve it and how will I make the solution compatable with my other solutions since they will have the ##\frac{1}{\epsilon}##
What does that integral mean? Usually when people write something like $\int dx^3$, that's shorthand for $\int dx \int dy \int dz$. I don't know what you mean by $\int d^4 l$

BiGyElLoWhAt
stevendaryl
Staff Emeritus
What does that integral mean? Usually when people write something like $\int dx^3$, that's shorthand for $\int dx \int dy \int dz$. I don't know what you mean by $\int d^4 l$
Hmm. Maybe I'm forgetting the convention. It looks like from Bill's post that $\int d^3 x$ is used, rather than $\int dx^3$. But in either case, it's not clear what $\int d^4 l \frac{1}{l^4}$ means. Is this supposed to be the same as $\int dl1 \int dl2 \int dl3 \int dl4 \frac{1}{\sqrt{l1^2 + l2^2 + l3^2 + l4^2}}$?

bhobba
Mentor
The issue is the metric defining l^2.

It some instances, and this is one, its clearer putting the dx etc in front.

Thanks
Bill

BiGyElLoWhAt
Gold Member
Could someone possibly shed some context on what's happening here? So l is in a curved space? I don't see how putting the differential(s) in front of the function clears things up. I'm apparently missing the point.

bhobba
Mentor
Could someone possibly shed some context on what's happening here? So l is in a curved space? I don't see how putting the differential(s) in front of the function clears things up. I'm apparently missing the point.

Thanks
Bill

BiGyElLoWhAt
Gold Member
I skimmed it, and it doesn't provide context. Are you referring to the F(k^2) generalization? It goes into details on how to solve it, but says nothing about the context. Thanks anyways.

bhobba
Mentor
I skimmed it, and it doesn't provide context. Are you referring to the F(k^2) generalization? It goes into details on how to solve it, but says nothing about the context.
They are integrals like you get in a math book. That's its context.

Thanks
Bill

fzero
Homework Helper
Gold Member
So I am calculating the renormalization group rquations for some exotic new particles and used dimensional regularization for all the calculations up to this point. Now I am look at the vertex corrections in the massless limit in which all external momentum are equal to 0. The issue here is I end up with an integral I don't know how to solve.

Its this integral
$$\int d^{4}l \frac{1}{l^4}$$
which you obviously can't solve using dimensional regularization. What other method can I use to solve it and how will I make the solution compatable with my other solutions since they will have the ##\frac{1}{\epsilon}##
You can treat this in dimensional regularization. Typically you would just say that this dimensionless integral gives zero by the Veltman rule, but we can be more specific. The interesting part is obviously

$$I = \int_0^\infty d\ell \ell^{d-5} = \left[ \frac{k^{d-4}}{d-4} \right]_0^\infty,$$

which is UV divergent if ##d>4## and IR divergent.if ##d<4##. We can isolate the divergences by introducing an intermediate scale ##\mu## and writing the integral as the sum

$$I = \int_0^\mu d\ell \ell^{d-5} + \int_\mu^\infty d\ell \ell^{d-5}.$$

The first term contains the IR divergence so we let ##d = 4 + \epsilon_\text{IR}##, with ##\epsilon_\text{IR}>0##. The 2nd term has the UV divergence, so there we let ##d=4-\epsilon_{UV}##. Putting these together, we find

$$I = \int_0^\mu d\ell \ell^{\epsilon_\text{IR}-1} + \int_\mu^\infty d\ell \ell^{-\epsilon_{UV}-1} = \frac{\mu^{\epsilon_\text{IR}}}{\epsilon_\text{IR}} + \frac{1}{\epsilon_\text{UV} \mu^{\epsilon_\text{UV}}} .$$

For small ##\epsilon_\text{IR},\epsilon_\text{UV}##,

$$I \approx \frac{1}{\epsilon_\text{IR}} + \log \mu + \frac{1}{\epsilon_\text{UV} } - \log \mu \approx \frac{1}{\epsilon_\text{IR}} + \frac{1}{\epsilon_\text{UV} }.$$

How to deal with these divergences now depends on the physical process under consideration. The UV divergence should be automatically canceled by the appropriate counterterm for the diagram being considered. For the IR divergence, we can simply include the subtraction in the renormalization scheme, e.g. MS. Otherwise we have to incorporate the diagram into a physical quantity. Then the possible IR divergences will cancel in the sum over the collection of diagrams that contribute to the physical quantity.